A wire that has a resistance of \(5.0 \Omega\) is passed through an extruder so
as to make it into a new wire three times as long as the original. What is the
new resistance?
Use \(R=\rho L / A\) to find the resistance of the new wire. To find \(\rho\), use
the original data for the wire. Let \(L_{0}\) and \(A_{0}\) be the initial length
and cross-sectional area, respectively. Then
$$
5.0 \Omega=\rho L_{0} / A_{0} \quad \text { or } \quad \rho=\left(A_{0} /
L_{0}\right)(5.0 \Omega)
$$
We were told that \(L=3 L_{0}\). To find \(A\) in terms of \(A_{0}\), note that the
volume of the wire cannot change. Hence,
$$
V_{0}=L_{0} A_{0} \quad \text { and } \quad V_{0}=L A
$$
from which
$$
L A=L_{0} A_{0} \quad \text { or } \quad
A=\left(\frac{L_{0}}{L_{0}}\right)\left(A_{0}\right)=\frac{A_{0}}{3}
$$
Therefore, \(\quad R=\frac{\rho L}{A}=\frac{\left(A_{0} / L_{0}\right)(5.0
\Omega)\left(3 L_{0}\right)}{A_{0} / 3}=9(5.0 \Omega)=45 \Omega\)